Bourbaki

Proposition 8

Tag 08O1, permanent. Cite it as [Bourbaki, Tag 08O1].

Integration, chapter III, § 3. Integrals of continuous vector-valued functions

— Suppose that E is Hausdorff; let μ\mu be a measure on X. The mapping f↦∫fdμf \mapsto \int f d\mu of K(X;E)\mathscr{K}(X;E) into E^\widehat{E} (No. 3, Cor. 1 of Prop. 7) is the unique continuous linear mapping Φ\Phi such that Φ(g⋅a)=μ(g)⋅a\Phi(g \cdot a) = \mu(g) \cdot a for every vector a∈Ea \in E and every function g∈K(X;C)g \in \mathscr{K}(X;\mathbf{C}).

To prove the continuity of the mapping f↦∫fdμf \mapsto \int f d\mu, it suffices to show that its restriction to K(X,K;E)\mathscr{K}(X,K;E) is continuous for every compact subset K of X (TVS, II, §4, No. 4, Prop. 5). We note that if the topology of E is defined by a family of semi-norms (qα)(q_\alpha), that of K(X,K;E)\mathscr{K}(X,K;E) is defined by the family of semi-norms

pα(f)=sup⁡x∈Kqα(f(x)). p_\alpha(f) = \sup_{x \in K} q_\alpha(f(x)) .

Now, let hh be a continuous mapping of X into [0,1][0,1], with compact support and such that h(x)=1h(x) = 1 on K; by No. 2, Prop. 6 we have, for every function f∈K(X,K;E)f \in \mathscr{K}(X,K;E),

qα(∫fdμ)=qα(∫hfdμ)≤∫h(x)qα(f(x))d∣μ∣(x)≤∣μ∣(h)⋅pα(f) q_\alpha \left( \int f d\mu \right) = q_\alpha \left( \int h f d\mu \right) \leq \int h(x) q_\alpha(f(x)) d|\mu|(x) \leq |\mu|(h) \cdot p_\alpha(f)

(the qαq_\alpha being extended by continuity to E^\widehat{E}), which proves the continuity of f↦∫fdμf \mapsto \int f d\mu. On the other hand, with the notations of the statement,

∫(g(x)⋅a)dμ(x)=μ(g)⋅a \int (g(x) \cdot a) d\mu(x) = \mu(g) \cdot a

by virtue of No. 1, Example 1 and Prop. 2 of No. 2 applied to the canonical injection C⋅a→E\mathbf{C} \cdot a \to E. Moreover, the subspace of K(X;E)\mathscr{K}(X;E) formed by the linear combinations ∑igi⋅ai\sum_i g_i \cdot a_i, where ai∈Ea_i \in E and gi∈K(X;C)g_i \in \mathscr{K}(X;\mathbf{C}), is dense in K(X;E)\mathscr{K}(X;E) (§1, No. 2, Prop. 5), which completes the proof.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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